# Math

posted by .

Determine the solution set to the equation: 2sin^2x+cosx-2=0

• Math -

I will assume your domain is 0 --- 360

2sin^2x+cosx-2=0
2(1- cos^2x) + cosx - 2 = 0
2 - 2cos^2x + cosx - 2 = 0
2cos^2x - cosx = 0
cosx(2cosx - 1) = 0
cosx = 0 or cosx = 1/2
x = 90º or 270º or x = 60º or x = 300º

these can be easily converted to radians

## Similar Questions

1. ### Math - Trig - Double Angles

Prove: sin2x / 1 - cos2x = cotx My Attempt: LS: = 2sinxcosx / - 1 - (1 - 2sin^2x) = 2sinxcosx / - 1 + 2sin^2x = cosx / sinx - 1 = cosx / sinx - 1/1 = cosx / sinx - sinx / sinx -- Prove: 2sin(x+y)sin(x-y) = cos2y - cos2x My Attempt: …
2. ### math

tanx+secx=2cosx (sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 = 2(1-sin^2x) 2sin^2x + sinx-1=0 (2sinx+1)(sinx-1)=0 x=30 x=270 but if i plug 270 back into the original equation …
3. ### trig

Find all solutions of the equation 2sin^2x-cosx=1 in the interval [0,2pi) x1= ?
4. ### pre-cal

show solution solve. 2sin^(2)x=2+cosx
5. ### math

solve each equation for 0=/<x=/<2pi sin^2x + 5sinx + 6 = 0?
6. ### Math

How many solutions does the equation cosx + 1/2 = 1 have for 0<x<2pi cosx+1/2=1 cosx+1=2 cosx=2-1 cosx=1 therefore 1 solution?
7. ### Math

Solve each equation for 0≤x<2π 1. (cosX)(cosX+1)=0 2. 2sin^2(X)-3sinX=2
8. ### Trigonometry

Solve for the following equation to the nearest degree if needed for 0 less than equal to x less than equal to 360. sin^2x=cosx 2sin^2x+cosx-cos^2x=0
9. ### Math

(cosx - sinx)^2 + (cosx + sinx)^2 = 2 Iam in this step: 2cos^2(x) + 2sin^2(x) = 2 How can I make the equation on the right equal 2
10. ### Math

(sinx - cosx)(sinx + cosx) = 2sin^2x -1 I need some tips on trigonometric identities. Why shouldn't I just turn (sinx + cosx) into 1 and would it still have the same identity?

More Similar Questions